What is the solution for (2+x^2)y^'-2xy=0 ?

The solution for (2+x^2)y^'-2xy=0 is y=2e^{c_{1}}+e^{c_{1}}x^2

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(2+x^2)y^'-2xy=0

(2 + x^{2}) y^{^{'}} - 2 x y = 0

y = 2 e^{c_{1}} + e^{c_{1}} x^{2}

Solution steps

(2 + x^{2}) y^{'} (x) - 2 x y = 0

Solve separable ODE:

y = 2 e^{c_{1}} + e^{c_{1}} x^{2}

y = 2 e^{c_{1}} + e^{c_{1}} x^{2}

y = 8 x - x^{2}

\frac{d}{d x} (x^{2} - 3 x y + y^{2})

\frac{d}{d x} (\frac{( x - 2 )}{x + 2})

y^{^{''}} + 30 y = 0

\int \frac{2 x}{2 x - 1} d x

What is the solution for (2+x^2)y^'-2xy=0 ?
The solution for (2+x^2)y^'-2xy=0 is y=2e^{c_{1}}+e^{c_{1}}x^2